 # Solve the third-order initial value problem below using the method of Laplace transforms y'''-2y"-21y'-18y=-18 y(0)=2 y'(0)=7 y"(0)=95 Khadija Wells 2020-11-09 Answered
Solve the third-order initial value problem below using the method of Laplace transforms
${y}^{‴}-2y"-21{y}^{\prime }-18y=-18$
$y\left(0\right)=2$
${y}^{\prime }\left(0\right)=7$
$y"\left(0\right)=95$
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Step 1
Taking Laplace Transform in both direction:
$L\left\{{y}^{‴}-2y"-21{y}^{\prime }-18y\right\}=L\left\{18\right\}$
$L\left\{{F}^{n}\left(t\right)\right\}={s}^{n}Y\left(s\right)-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y"\left(0\right)$
$L\left\{{y}^{‴}\right\}={s}^{3}Y\left(s\right)-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left\{y"\right\}={s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)$
$L\left\{{y}^{\prime }\right\}=sY\left(s\right)-y\left(0\right)$
$L\left\{y\right\}=Y\left(s\right)$
$⇒\left[{s}^{3}Y\left(s\right)-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-{y}^{\prime }\left(0\right)\right]-2\left[{s}^{2}Y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]-21\left[sY\left(s\right)-y\left(0\right)\right]-18\left[Y\left(s\right)\right]=L\left\{-18\right\}$
$⇒\left[{s}^{3}Y\left(s\right)-{s}^{2}\left(2\right)-7s-95\right]-2\left[{s}^{2}Y\left(s\right)-s\left(2\right)-7\right]-21\left[sY\left(s\right)-2\right]-18\left[Y\left(s\right)\right]=-\frac{18}{s}$
$⇒{s}^{3}-2{s}^{2}-21s-18Y\left(s\right)-2{s}^{2}-7s-95+4s+14+42=-\frac{18}{s}$
$⇒Y\left(s\right)=\frac{2{s}^{2}+3s-\frac{18}{s}+39}{{s}^{3}-2{s}^{2}-21s-18}=\frac{2{s}^{3}+3{s}^{2}-18+39s}{s\left(s+1\right)\left(s+3\right)\left(s-6\right)}$
Step 2
find the factor:
$Y\left(s\right)=\frac{A}{s}+\frac{B}{\left(s+1\right)}+\frac{C}{\left(s+3\right)}+\frac{D}{\left(s-6\right)}$
$Y\left(s\right)=\frac{2{s}^{3}+3{s}^{2}-18+39s}{s\left(s+1\right)\left(s+3\right)\left(s-6\right)}$
$\frac{A}{s}+\frac{B}{\left(s+1\right)}+\frac{C}{\left(s+3\right)}+\frac{D}{\left(s-6\right)}=\frac{\frac{2\left(0{\right)}^{3}+3\left(0{\right)}^{2}-18+39\left(0\right)}{\left(1\right)\left(3\right)\left(-6\right)}}{s}+\frac{\frac{2\left(-1{\right)}^{3}+3\left(-1{\right)}^{2}-18+39\left(-1\right)}{\left(-1\right)\left(-1+3\right)\left(-1-6\right)}}{\left(s+1\right)}+\frac{\frac{2\left(-3{\right)}^{3}+3\left(-3{\right)}^{2}}{}}{}$